Question Number 187790 by cherokeesay last updated on 21/Feb/23
Answered by mr W last updated on 21/Feb/23
Commented by mr W last updated on 22/Feb/23
$$\left(\mathrm{2}{a}−\mathrm{2}\right)^{\mathrm{2}} +\left[\left(\sqrt{\mathrm{5}}−\mathrm{1}\right){a}\right]^{\mathrm{2}} =\left(\mathrm{2}{a}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{4}}{\mathrm{3}−\sqrt{\mathrm{5}}}=\mathrm{3}+\sqrt{\mathrm{5}} \\ $$$${area}\:{of}\:{square}\:=\left(\mathrm{2}{a}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$\:\:=\mathrm{56}+\mathrm{24}\sqrt{\mathrm{5}}\approx\mathrm{109}.\mathrm{666} \\ $$
Commented by cherokeesay last updated on 22/Feb/23
$${so}\:{nice}\:{thank}\:{you}\:{sir}. \\ $$
Answered by HeferH last updated on 22/Feb/23
$${let}\:\mathrm{2}{x}\:{be}\:{the}\:{side}\:{of}\:{the}\:{square}: \\ $$$$\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} +\left({x}\sqrt{\mathrm{5}}−{x}\right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} +\left(\mathrm{2}{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}−\mathrm{8}{x}\:+\:\mathrm{5}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} \sqrt{\mathrm{5}}\:=\:\mathrm{4}\:+\:\mathrm{4}{x}^{\mathrm{2}} \\ $$$$\:\mathrm{6}{x}^{\mathrm{2}} −\mathrm{8}{x}−\mathrm{2}{x}^{\mathrm{2}} \sqrt{\mathrm{5}}\:=\:\mathrm{0} \\ $$$$\:\mathrm{6}{x}\:−\mathrm{8}\:−\mathrm{2}{x}\sqrt{\mathrm{5}}\:=\:\mathrm{0} \\ $$$$\:\mathrm{3}{x}−{x}\sqrt{\mathrm{5}}\:=\:\mathrm{4} \\ $$$$\:{x}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)\:=\:\mathrm{4} \\ $$$$\:{x}\:=\:\frac{\mathrm{4}}{\mathrm{3}−\sqrt{\mathrm{5}}}\:;\:\mathrm{4}{x}^{\mathrm{2}} =\mathrm{4}\left[\:\frac{\mathrm{4}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)}{\mathrm{9}−\mathrm{5}}\right]^{\mathrm{2}} \:=\:\mathrm{4}\left(\mathrm{14}+\mathrm{6}\sqrt{\mathrm{5}}\right)\:{u}^{\mathrm{2}} = \\ $$$$\:\mathrm{56}+\mathrm{24}\sqrt{\mathrm{5}}{u}^{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 22/Feb/23
$${perfect}\:! \\ $$$${thank}\:{you}\:{sir}. \\ $$